$ \int_{-1}^0 \int_0^{3 - 3x^2} dy \, dx$ Switch the bounds of the double integral. Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^1 \int_{-\sqrt{1-y/3}}^0 dx \, dy$ (Choice B) B $ \int_0^3 \int_{\sqrt{1-y/3}}^0 dx \, dy$ (Choice C) C $ \int_0^1 \int_{\sqrt{1-y/3}}^0 dx \, dy$ (Choice D) D $ \int_0^3 \int_{-\sqrt{1-y/3}}^0 dx \, dy$
Solution: The first step whenever we want to switch bounds is to sketch the region of integration that we're given. Here, we see $-1 < x < 0$ and $0 < y < 3 - 3x^2$. Therefore: ${1}$ ${\llap{-}1}$ ${\llap{-}2}$ ${1}$ ${2}$ ${3}$ $y$ $x$ Because we're switching bounds to $dx \, dy$, we need to start with numeric bounds for $y$. We see that $0 < y < 3$. Then we can define $x$ in terms of $y$. Thus, $-\sqrt{1-y/3} < x < 0$. We want to pay attention especially to how this $x$ bound works at the edge of the $y$ bound. For example, at $y = 0$, the $x$ bound makes $x = -1$ as expected. In conclusion, the double integral after switching bounds is: $ \int_0^3 \int_{-\sqrt{1-y/3}}^0 dx \, dy$